-16t^2+340t+8=0

Solution for -16t^2+340t+8=0 equation:

-16t^2+340t+8=0

a = -16; b = 340; c = +8;
Δ = b2-4ac
Δ = 3402-4·(-16)·8
Δ = 116112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{116112}=\sqrt{16*7257}=\sqrt{16}*\sqrt{7257}=4\sqrt{7257}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(340)-4\sqrt{7257}}{2*-16}=\frac{-340-4\sqrt{7257}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+4\sqrt{7257}}{2*-16}=\frac{-340+4\sqrt{7257}}{-32}
$